Let $R\to S$ be an extension of possibly non-commutative rings. I am interested in the relationship between $R\to S$ being Frobenius and it being separable.

If it is a Frobenius extension, then there is certain canonical ideal $\Gamma\subseteq Z(S)$ that (among other things) has the property that Frobenius extensions with $\Gamma=Z(S)$ are automatically separable.

If $S$ is not an arbitrary extension of some arbitrary $R$ but a $K$-algebra for some field $K$, then a reverse holds: Every separable $K$-extension is automatically a Frobenius algebra (even a symmetric algebra) with $\Gamma=Z(S)$.

My question is, whether this holds more generally:

Is every separable extension $R\to S$ automatically a Frobenius extension? (modulo some technical conditions like $S$ being projective over $R$)

Just for completeness sake, a quick recap of why I think this should be true.

$R\to S$ is Frobenius if there exists a $R$-$R$-linear form $\tau: S\to S$ and $x_i, y_i\in S$ such that $s=\sum_i \tau(sx_i)y_i$.

(The Casimir element $\Omega:=\sum_i x_i\otimes y_i\in S\otimes S$ is independent of the choice of $x_i, y_i$ and only mildly dependend of the choice of $\tau$.)

Every Frobenius extension induces the Gaschütz-Ikeda-operator; a natural transformation $I: Hom_R(Res(V),Res(W)) \to Hom_S(V,W)$ by $$I(f) := v \mapsto \sum_i x_i f(y_i v)$$ (definition is independent of choice of $x_i,y_i$ and only mildly dependend on the choice of $\tau$)

The image $I(End(V))$ measures how much $V$ fails to be relatively projective, i.e. $I(End(V)) = End(V) \iff V$ is relatively projective.

In the special case $V=S$ we get an ideal $\Gamma:=I(C_S(R)) = \{\sum_i a_i c b_i \mid c\in C_S(R)\}\subseteq Z(S)$ such that $\Gamma=Z(S)$ iff $S$ is relatively projective iff all $S$-modules are relatively projective.

Now for separable extensions, the situation is very similar: $R\to S$ is separable iff there is an $e=\sum_i a_i\otimes b_i\in S\otimes_R S$ such that $\sum_i a_i b_i = 1$ and $\forall s\in S: sx=xs$. Every choice of such an $e$ also defines a natural transformation $M: Hom_R(Res(V),Res(W)) \to Hom_S(V,W)$ by setting $$I(f) := v\mapsto \sum_i a_i f(b_i v)$$ and one can show that this an idempotent projection onto the subspace $Hom_S(V,W)$ so that all $S$-modules are automatically relatively projective. In fact $Hom_S(V,W)$ being a direct summand of $Hom_R(V,W)$ in a natural way is equivalent to $R\to S$ being separable.

That way Frobenius+$\Gamma=Z(S)$ implies separable and the proof shows that the element $e$ can be chosen to be of the form $\sum x_i\otimes cy_i$ for some $c\in C_S(R)$ so that the separability element $e$ and the Casimir element $\Omega$ are closely related to each other if not identical.

Given the properties of both operators $I$, an equivalent question to ask would be:

Is $R\to S$ being Frobenius equivalent (again modulo technicalities) to the existence of a natural transformation $I: Hom_R(Res(-),Res(-)) \to Hom_S(-,-)$ such that $id_V \in im(I) \iff V$ is relatively projective?

  • $\begingroup$ I'm not sure if it has exactly what you want, but have you looked at: Nakane, K., Note on separable extensions of rings, Sci. Rep. Tokyo Kyoiku Daigaku, Sect. A 10, 142-145 (1969). ZBL0207.34501? $\endgroup$ – Jeremy Rickard Jun 8 '17 at 10:15

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